Evaluate the indefinite integral given below.
∫
(
sin
3
(
x
)
−
3
sin
2
(
x
)
)
(
sin
4
(
x
)
−
4
sin
3
(
x
)
)
8
cos
(
x
)
d
x

To evaluate the indefinite integral:

\[
\int \left( \sin^3(x) - 3\sin^2(x) \right) \left( \sin^4(x) - 4\sin^3(x) \right) 8\cos(x) \, dx
\]

### Step 1: Substitution
Let:
\[
u = \sin(x)
\]
Then:
\[
du = \cos(x) \, dx
\]

The integral becomes:
\[
\int \left( u^3 - 3u^2 \right) \left( u^4 - 4u^3 \right) 8 \, du
\]

### Step 2: Expand the product
Expand \( \left( u^3 - 3u^2 \right) \left( u^4 - 4u^3 \right) \):
\[
(u^3 - 3u^2)(u^4 - 4u^3) = u^7 - 4u^6 - 3u^6 + 12u^5 = u^7 - 7u^6 + 12u^5
\]

The integral becomes:
\[
\int 8 \left( u^7 - 7u^6 + 12u^5 \right) du
\]

### Step 3: Simplify and integrate term by term
Distribute the constant \(8\) and integrate:
\[
\int 8(u^7 - 7u^6 + 12u^5) \, du = 8 \int u^7 \, du - 8 \int 7u^6 \, du + 8 \int 12u^5 \, du
\]

Integrate each term:
\[
8 \int u^7 \, du = \frac{8u^8}{8} = u^8, \quad
8 \int 7u^6 \, du = 8 \cdot \frac{7u^7}{7} = 8u^7, \quad
8 \int 12u^5 \, du = 8 \cdot \frac{12u^6}{6} = 16u^6
\]

Thus:
\[
u^8 - 8u^7 + 16u^6 + C
\]

### Step 4: Back-substitute \(u = \sin(x)\)
Substitute back \(u = \sin(x)\):
\[
\sin^8(x) - 8\sin^7(x) + 16\sin^6(x) + C
\]

### Final Answer:
\[
\boxed{\sin^8(x) - 8\sin^7(x) + 16\sin^6(x) + C}
\]

Let me know if you have questions!  
### Follow-Up Questions:
1. How is substitution used to simplify trigonometric integrals?
2. What are common strategies for expanding and integrating products of polynomials?
3. Why is \(du = \cos(x) \, dx\) necessary for substitution in trigonometric integrals?
4. How can you verify the result by differentiation?
5. What role does symmetry play in definite integrals of trigonometric functions?

### Tip:
Always expand polynomials carefully and verify term-by-term integration to

Evaluate the indefinite integral given below. ∫(sin³(x) − 3sin²(x))(sin⁴(x) − 4sin³(x))8cos(x)dx

Explore the step-by-step solution for evaluating the integral ∫(sin³(x) − 3sin²(x))(sin⁴(x) − 4sin³(x))8cos(x)dx using trigonometric substitution and polynomial expansion. This advanced calculus problem demonstrates the substitution method, power rule for integration, and careful back-substitution techniques. Suitable for Grades 11-12 and early college students.

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